1. **State the problem:** We want to find the limit as $n \to \infty$ of the sequence $$S_n = \sin \left( \frac{(n+1) \pi}{12n+11} \right).$$ 2. **Analyze the argument inside the sine:** Consider the argument of the sine function: $$\theta_n = \frac{(n+1) \pi}{12n+11}.$$ 3. **Simplify the fraction inside the argument:** Divide numerator and denominator by $n$: $$\theta_n = \frac{(n+1) \pi}{12n+11} = \frac{\pi \left(1 + \frac{1}{n}\right)}{12 + \frac{11}{n}}.$$ 4. **Take the limit of the argument as $n \to \infty$:** As $n \to \infty$, $\frac{1}{n} \to 0$, so $$\lim_{n \to \infty} \theta_n = \frac{\pi (1 + 0)}{12 + 0} = \frac{\pi}{12}.$$ 5. **Use continuity of sine to find the limit of $S_n$:** Since sine is continuous, $$\lim_{n \to \infty} S_n = \sin \left( \lim_{n \to \infty} \theta_n \right) = \sin \left( \frac{\pi}{12} \right).$$ 6. **Evaluate $\sin(\pi/12)$:** Recall that $\pi/12 = 15^\circ$, and $$\sin \frac{\pi}{12} = \sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4} \approx 0.2588.$$ **Final answer:** $$\boxed{0.259}$$ (rounded to three decimal places)